Question

15. In a class of 10 boys and 12 girls, four students are to be chosen to serve on a committee. What is the probability that: a. All 4 members of the committee will be girls? b. All 4 members of the committee will be boys? C. There will be at least one girl on the committee? 1

Answer 1

A)

We will solve using:

`_(22)C_4=(22\cdot21\cdot20\cdot19)/(4\cdot3\cdot2)=7315`

The number of ways to choose 4 girls from 12 girls and 0 boys from 10 boys is:

`_(10)C_0\cdot_(12)C_4=1\cdot(12!)/(8!\cdot4!)=(12\cdot11\cdot10\cdot9)/(4\cdot3\cdot2)=495`

So, the probability() of choosing 4 just 4 girls is:

`P=(_(10)C_0\cdot_(12)C_4)/(_(22)C_4)\Rightarrow P=(495)/(7315)\Rightarrow P=(9)/(133)\Rightarrow P\approx0.068`

So, that is the probability to get just 4 girls on the committee.

B)

For the committee to be just 4 boys is found as follows:

`_(22)C_4=(22\cdot21\cdot20\cdot19)/(4\cdot3\cdot2)=7315`

And the number of ways to choose 4 boys from 10 boys and 0 girls from 12 girls is:

`_(10)C_4\cdot_(12)C_0=(10\cdot9\cdot8\cdot7)/(4\cdot3\cdot2)=210`

So, the probability to get just 4 boys on the committee is:

`P=(_(10)C_4\cdot_(12)C_0)/(_(22)C_4)\Rightarrow P=(210)/(7315)\Rightarrow P=(6)/(209)\Rightarrow P\approx0.028`

C)

For the committee to have at least one girl is:

`_(22)C_4=(22\cdot21\cdot20\cdot19)/(4\cdot3\cdot2)=7315`

No. of ways to selecting at least 1 girl:

22C4 - 10C4 = 7315 - 210 = 7105

Now, we calculate the probability:

`P=(7105)/(7315)\Rightarrow P=(203)/(209)\Rightarrow P\approx0.971`